I am interested in figures of elementary geometry that can be used to illustrate the Hausdorff distance. You don't even have to know what it is. Here's what it looks like in geometrical terms:
Let $T=ABC$ be a triangle (with its interior) and let $FrT=Fr(ABC)=AB\cup BC\cup CA$ be the boundary of $ABC$. I'm looking for maximum of $$\{d(M,\text{Fr}T),M\in T\}$$
If $M\notin $ bissector of $\widehat{A}$, suppose that $M$ is closer to $AB$ than to $AC$, then, with the notations in the figure, $d(M,AB)=MI<IG=GH=d(G, AB\cup AC)$
So the farthest point from $FrT$ is necessarily on the bisector of $\widehat{A}$.
Similarly, it will be on the bisector of $\widehat{B}$.
So that's $D$, the center of the circle in $ABC$. And the distance we are looking for is then $r$, the radius of the inscribed circle.
I am not absolutely convinced by what I have written. So I would appreciate the time you would take to proofread.
There seems to be a problem in your reasoning. Suppose $G$ is almost near $BC$. Then $d(M, FrT)$ is greater than $d(G,FrT)$. Though it’s true that $d (M, AB)<d(G,AB\cup AC)$.
One can reason as follows. Suppose $M$ is not $D$. Then it is (at least) in one of the three triangles: $ABD$, $BCD$, $CAD$. Let it be the triangle $ABD$. But then $$d(M,AB)<d(D,AB).$$ And since $$d(M,FrT)=\min(d(M,AB), d(M,BC), d(M,CA))\le d(M,AB)$$ and $$d(D,AB)=d(D,FrT),$$ we have $$d(M,FrT)< d(D,FrT).$$